Suppose a, b, c are three distinct real numbe | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a)

Given,

$P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$

$P(a)=1+0+0=1$

$P(b)=0+1+0=1$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a)

Given,

$P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$

$P(a)=1+0+0=1$

$P(b)=0+1+0=1$

$P(c)=0+0+1=1$

$P(x)$ is a polynomial of degree atmost $2$ and also attains same value i.e.$1$ for distinct values of $x$ (i.e. $a, b, c$ ).

$	herefore P(x)$ is an identity with only value equal to $1$ for all $R$.

$	herefore \frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}=1$

Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes

Similar Questions

The roots of the equation ${x^4} - 4{x^3} + 6{x^2} - 4x + 1 = 0$ are

If $x$ is real, the expression $\frac{{x + 2}}{{2{x^2} + 3x + 6}}$ takes all value in the interval

If $x$ is real , the maximum value of $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ is

The solution of the equation $2{x^2} + 3x - 9 \le 0$ is given by

The number of real solutions of the equation $|{x^2} + 4x + 3| + 2x + 5 = 0 $are